# On transmission coefficients in QM case -checking my reasoning

1. Apr 20, 2014

### Emspak

1. The problem statement, all variables and given/known data

The task is to calculate the transmission probability when E<V0 given that the step potential barrier is of infinite length.

2. Relevant equations

So far it seems simple enough. For region I, where x<0, (we're assuming, as usual, that the particles/ waves are coming from the left) I had

$\phi''_I = \frac{2mE}{\hbar^2}\phi$ and $\phi''_{II} = \frac{2m(E-V_0)}{\hbar^2}\phi$

and we will let $\frac{2mE}{\hbar^2} = k_1^2$ and $\frac{2m(E-V_0)}{\hbar^2} = k_2^2$

3. The attempt at a solution

SO I go through the whole process of solving the DQ.

$\phi_I = Ae^{ik_1x} + Be^{-ik_1x}$
$\phi_{II} = Ce^{ik_2x} + De^{-ik_2x}$

$\phi_I' = ik_1Ae^{ik_1x} -ik_1Be^{-ik_1x}$
$\phi_{II}' = ik_2Ce^{ik_2x} -ik_2De^{-ik_2x}$

The two first derivatives are also equal at x=0, which sets another boundary condition.

D=0 here because the particles are all moving from the left to right and there isn't anything coming from the right.

That gets me

$ik_1A -ik_1B = ik_2C \rightarrow k_1(-A + B) = k_2C \rightarrow \frac{C}{A} = \frac{2k_1}{k_1+k_2}$

and the transmission probability is
$\frac{|C|^2}{|A|^2} = \frac{|2k_1|^2}{|k_1+k_2|^2}$

So far so good, I think. But then I noticed something. When I plugged in my k1 and k2 this happened:

$$\frac{|2k_1|^2}{|k_1+k_2|^2}=\frac{ \frac{8mE}{\hbar^2}}{\frac{2mE}{\hbar^2} +2 \left(\frac{\sqrt{2mE}}{\hbar} \right) \left(\frac{\sqrt{2m(E-V_0)}}{\hbar}\right)+\frac{2m(E-V_0)}{\hbar^2}}=\frac{4E}{2E+2\sqrt{2E}\sqrt{(E-V_0)}-V_0}$$

Anyhow I was just curious if I did this right because what was interesting was that you get an imaginary component in there but I could make it into a complex conjugate. (At least the first two terms). And I was interested in whether I should do that to get a real coefficient. I suppose it's just algebra, but I wanted to make sure I did this right from the get go.

2. Apr 21, 2014

### vela

Staff Emeritus
One problem is that your $k_2$ is imaginary, so $\phi_{II}$ doesn't represent propagating waves. If instead you define $k_2^2 = \frac{2m(V_0-E)}{h^2}$ so that $k_2$ is real, you can see that the solution in region II is $\phi_{II} = Ce^{-k_2x} + De^{k_2x}$. For the wave function to remain bounded as $x \to \infty$, you have to have D=0.

You actually have to compare the current densities, given by
$$j = \frac{\hbar}{2mi}\left(\psi^*\frac{\partial \psi}{\partial x}-\psi\frac{\partial \psi^*}{\partial x}\right).$$ The transmission and reflection coefficients are then given by $T = \left\lvert \frac{j_\text{trans}}{j_\text{inc}}\right\rvert$ and $R = \left\lvert \frac{j_\text{refl}}{j_\text{inc}}\right\rvert$. If you work out the reflection coefficient, you should be able to show that it's given by $R = \lvert B/A \rvert^2 = 1$. The transmission coefficient must therefore vanish. This should make sense physically since the particle must always be reflected because E < V0 and the barrier is infinitely long.

Moreover, the transmission coefficient isn't generally given by $T = \lvert C/A \rvert^2$. This only holds if the potential V(x) is the same in the two regions. In this problem, besides the fact that in region II, you don't have a propagating wave, you also have that V(x) is different for x<0 and x>0.

Though this isn't the correct way to get the transmission coefficient, I thought I should still note that because of the way you defined $k_2$, you can't say that $\lvert k_1 + k_2 \rvert^2 = (k_1 + k_2)^2$ because $k_2$ is imaginary. Remember that $\lvert z \rvert^2 = z z^*$, which will always yield a real result.

Last edited: Apr 21, 2014
3. Apr 21, 2014

### Emspak

OK, so I just want to make sure, did I do the problem right, though, at lest up to the last part? I was told that $\frac{|C|^2}{|A|^2}+\frac{|B|^2}{|A|^2}=1$

and given what you told me about the actual calculation of transmission and reflection coefficients - the problem says a single particle, does that still apply? There isn't flux there in that case, it's just one, so is the transmission coefficient C/A at that point?

4. Apr 21, 2014

### vela

Staff Emeritus
No, you didn't do it correctly. That was the main point of my post!

5. Apr 21, 2014

### Emspak

OK, r-doing this a bit I tried re-defining some constants this way (I also noticed that I might have messed up a negative sign):

The time independent schrodinger is:

$$\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2}= (E-V)\psi(x)$$

(note the minus sign at the front) So when I set up my solution I should define my constants as:

$k_1^2 = \frac{-2mE}{\hbar^2}E$ and $k_1^2 = \frac{-2m(E-V}{\hbar^2}E$

sort of as before. But this time the solution set-up should be:

$Ae^{ik_1x}+Be^{ik_1x}=\phi_{I}$ and $Ce^{-ik_2x}+De^{ik_2x}=\phi_{I}$ and D goes away because there is no wave coming from the left.

Note that this time I made my exponent negative for C so it decays away past x=0 as one might expect. When I go through the whole process again this time I got (once I set up my boundary conditions)

$2ik_1A = -k_2(A+B) + ik_1(A+B)$

and doing the algebra for B/A

$\frac{B}{A} = \frac{ik_1-k_2)}{ik_1+k_2}$

and when I square that I get

$\left(\frac{B}{A}\right)^2 = \frac{-k_1^2+2k_1k_2+k_2^2}{-k_1^2-2k_1K_2+k_2^2}$

and since k2 is an imaginary number and it is multiplied in the second terms of the quadratic by i, then it's all real. I ended up with, for instance in the nominator

$\frac{-2m}{\hbar^2}+i\sqrt{\frac{2m}{\hbar^2}}2i\sqrt{\frac{2m(V-E)}{\hbar^2}}+\frac{2m(E-V)}{\hbar^2}= \frac{2m}{\hbar^2}-2\sqrt{\frac{2m}{\hbar^2}}\sqrt{\frac{2m(V-E)}{\hbar^2}}+\frac{2m(E-V)}{\hbar^2}=\frac{-2m}{\hbar^2}-2\sqrt{\frac{4m^2(E-V)}{\hbar^4}}+\frac{2m(E-V)}{\hbar^2}$

taking out -2m/h-bar squared :

$\frac{-2m}{\hbar^2}(1+2\sqrt{(V-E)}-(E-V))$
the denominator gets something similar, except it is

$\frac{-2m}{\hbar^2}(1-2\sqrt{(V-E)}-(E-V))$

so, did I get it this time :-) ?

6. Apr 21, 2014

### Emspak

I also just realized that since the nominator and denominator of B/A can be multiplied by the complex conjugates to get the absolute value squared I didn't need to do so much algebra.

7. Apr 21, 2014

### vela

Staff Emeritus
No, you didn't get it. I already told you R=1, and your expression doesn't give that. Did you really need me to tell you that?

8. Apr 21, 2014

### Emspak

Doing it again, I am getting R=1, and T=0, but I guess what's hanging me up is that if I calculate T separately I end up with nonzero values. It's weird. Now, maybe that's just a function of QM and the fact that Cexp(-k2)x is a real function and so you get the evanescent wave. I tried your method BTW and got a zero for R. I guess what's not satisfying me is that I feel like the transmission coefficient formula you have there (and when I go over it in the text) seems a bit, I dunno, arbitrary. But I will freely admit I mightn't have the ability/background to do a formal proof. I don't recall seeing the T formula you have in the class, is all. And it isn't in my notes, so I suspect there's some other method they want us to use.

9. Apr 23, 2014

### vela

Staff Emeritus
I don't know what you did, but "my way" and "your way" of calculating R are the same, so you shouldn't be getting different answers. Your method for calculating T, however, is incorrect for the reasons I said above.

I'm not sure why you think the formula I gave is any more arbitrary than someone telling you (incorrectly) that $T = |C/A|^2$. I'm sure your book discusses how to calculate T as does the following Wikipedia article.

http://en.wikipedia.org/wiki/Probability_current

If you want to avoid this kind of calculation, you can always calculate R and find T=1-R. The point of this problem may very well have been to get you to figure out why the method you used doesn't work.

10. Apr 23, 2014

### Emspak

I eventually figured out what the problem was (I think) -- the issue I was having was that I thought there was supposed to be a non-zero transmission coefficient, since this being QM there's a small probability it will get through. But then I realized that you just need to mention the evanescent wave (which decays exponentially) and that's ok, at leas in terms of this problem. Sorry to waste your time -- I was totally screwed up in how I understood the math.